A338939 a(n) is the number of partitions n = a + b such that a*b is a perfect square.

0, 1, 0, 1, 1, 1, 0, 1, 0, 3, 0, 1, 1, 1, 1, 1, 1, 1, 0, 3, 0, 1, 0, 1, 2, 3, 0, 1, 1, 3, 0, 1, 0, 3, 1, 1, 1, 1, 1, 3, 1, 1, 0, 1, 1, 1, 0, 1, 0, 5, 1, 3, 1, 1, 1, 1, 0, 3, 0, 3, 1, 1, 0, 1, 4, 1, 0, 3, 0, 3, 0, 1, 1, 3, 2, 1, 0, 3, 0, 3, 0, 3, 0, 1, 4, 1, 1, 1, 1, 3, 1, 1, 0, 1, 1, 1, 1, 1, 0, 5

Offset

1,10

Comments

Number of ways to regroup the unit squares of a rectangle with semiperimeter n into a square.

a(n) > 0 for n in A337140.

Links

David A. Corneth, Table of n, a(n) for n = 1..10000

Formula

Let n = 2^t * p_1^a_1 * p_2^a_2 * ... * p_r^a_r * q_1^b_1 * q_2^b_2 *...* q_s^b_s with t >= 0, a_i >= 0 for i = 1..r, where p_i == 1 (mod 4) and q_j == -1 (mod 4) for j = 1..s. Further, let A = (2a_1 + 1)*(2a_2 + 1)*...*(2a_r + 1). Then a(n) = (A-1)/2 for odd n and a(n) = A for even n.

a(n) = A338940(n) / A115878(n).

Example

n = 10 = 1 + 9 = 2 + 8 = 5 + 5 with 1*9 = 3^2 and 2*8 = 4^2 and 5*5 = 5^2. Then a(10) = 3. Also 10 = 2^1 * 5^1. So t=1, a_1=1 and a(n) = A = 2*1+1 = 3.

Mathematica

a[n_] := Count[IntegerPartitions[n, {2}], _?(IntegerQ @ Sqrt[Times @@ #] &)]; Array[a, 100] (* Amiram Eldar, Nov 22 2020 *)

Prog

(PARI) a(n)=my(c=0); for(i=1, n-1, if((2*i<=n)&&issquare(i*(n-i)), c++)); c
for(n=1, 100, print1(a(n), ", ")) \\ Derek Orr, Nov 18 2020
(PARI) a(n) = sum(i=1, n-1, (2*i<=n) && issquare(i*(n-i))); \\ Michel Marcus, Dec 21 2020
(PARI) first(n) = {my(res = vector(n)); for(i = 1, n, d = divisors(i^2); for(i = (#d + 1)\2, #d, c = d[i] + d[#d + 1 - i]; if(c <= n, res[c]++ , next(2) ) ) ); res } \\ David A. Corneth, Dec 21 2020

Crossrefs

Cf. A115878, A337140, A338940.

Sequence in context: A343858 A065714 A110700 * A292150 A181875 A051908

Adjacent sequences: A338936 A338937 A338938 * A338940 A338941 A338942

Keyword

nonn

Author

Hein van Winkel, Nov 16 2020

Status

approved